worksheet 6: section 3.2 and 3.3: differentiable functions derivatives of combinations of functions\ninstruct…

1.

Explanation:

Step1: Recall derivative property

The tangent line is horizontal when $f'(x)=0$. First, find the derivative of $f(x)=x^{3}-4x^{2}-16x + 3$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=3x^{2}-8x-16$.

Step2: Solve the quadratic equation

Set $f'(x)=0$, so $3x^{2}-8x - 16=0$. Factor the quadratic equation: $3x^{2}-8x - 16=(3x + 4)(x - 4)=0$.

Step3: Find the x - coordinates

Set each factor equal to zero: If $3x+4 = 0$, then $x=-\frac{4}{3}$; if $x - 4=0$, then $x = 4$.

Answer:

$x=-\frac{4}{3},4$

2.

Explanation:

Step1: Recall the velocity - displacement relationship

The velocity $v(t)$ is the derivative of the displacement function $s(t)$. Given $s(t)=2t^{3}+5t + 6$, using the power - rule $(x^n)'=nx^{n - 1}$, we get $v(t)=s'(t)=6t^{2}+5$.

Step2: Evaluate the velocity at $t = 1$

Substitute $t = 1$ into $v(t)$. So $v(1)=6(1)^{2}+5=6 + 5=11$.

Answer:

$11$ feet per second

3.

Explanation:

Step1: Find the derivative of $h(x)$

Given $h(x)=\sqrt{x}=x^{\frac{1}{2}}$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $h'(x)=\frac{1}{2\sqrt{x}}$.

Step2: Find the slope of the given line

The line $4y=x + 4$ can be written in slope - intercept form $y=\frac{1}{4}x + 1$, so its slope $m=\frac{1}{4}$.

Step3: Set the derivative equal to the slope

Set $h'(x)=\frac{1}{4}$, so $\frac{1}{2\sqrt{x}}=\frac{1}{4}$. Cross - multiply to get $2\sqrt{x}=4$, then $\sqrt{x}=2$, and $x = 4$.

Step4: Find the y - coordinate of the point of tangency

Substitute $x = 4$ into $h(x)$. Since $h(4)=\sqrt{4}=2$.

Answer:

The point of tangency is $(4,2)$

4.

Explanation:

(a)

Since the graph of $y = f(x)=ax^{2}+bx + c$ passes through the point $(0,2)$, substitute $x = 0$ and $y = 2$ into $f(x)$. We get $f(0)=a(0)^{2}+b(0)+c=2$, so $c = 2$.

(b)

First, find the derivative $f'(x)=2ax + b$. Then $f'(0)=b$. Since the line $y = 3x+2$ is tangent to the graph of $f$ at $(0,2)$, the slope of the tangent line is $3$, so $b = 3$.

(c)

Since $f(x)=ax^{2}+3x + 2$ and the graph of $f$ passes through the point $(-1,6)$. Substitute $x=-1$ and $y = 6$ into $f(x)$: $a(-1)^{2}+3(-1)+2=6$. Simplify to $a-3 + 2=6$, then $a=7$.

Answer:

(a) $c = 2$ (b) $b = 3$ (c) $a = 7$

5.

Explanation:

Step1: Use the quotient rule

Given $f(x)=\frac{x + 1}{x - 1}$, by the quotient rule $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^{2}}$, where $u=x + 1$, $u'=1$, $v=x - 1$, $v'=1$. Then $f'(x)=\frac{1\cdot(x - 1)-(x + 1)\cdot1}{(x - 1)^{2}}=\frac{x-1-x - 1}{(x - 1)^{2}}=-\frac{2}{(x - 1)^{2}}$.

Step2: Find the slope at $x = 3$

Substitute $x = 3$ into $f'(x)$: $f'(3)=-\frac{2}{(3 - 1)^{2}}=-\frac{2}{4}=-\frac{1}{2}$.

Step3: Find the y - coordinate at $x = 3$

Substitute $x = 3$ into $f(x)$: $f(3)=\frac{3 + 1}{3 - 1}=\frac{4}{2}=2$.

Step4: Use the point - slope form

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(3,2)$ and $m=-\frac{1}{2}$. So $y-2=-\frac{1}{2}(x - 3)$, which simplifies to $y=-\frac{1}{2}x+\frac{3}{2}+2=-\frac{1}{2}x+\frac{7}{2}$, or $x + 2y=7$.

Answer:

$x + 2y=7$

6.

Explanation:

Step1: Use the product rule

Given $g(x)=xf(x)$, by the product rule $(uv)'=u'v+uv'$, where $u=x$, $u'=1$, $v=f(x)$, $v'=f'(x)$. Then $g'(x)=f(x)+xf'(x)$.

Step2: Evaluate $g(3)$ and $g'(3)$

We know that $f(3)=4$ and $f'(3)=-2$. So $g(3)=3f(3)=3\times4 = 12$, and $g'(3)=f(3)+3f'(3)=4+3\times(-2)=4-6=-2$.

Step3: Use the point - slope form

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(3,12)$ and $m=-2$. So $y-12=-2(x - 3)$, which simplifies to $y=-2x+6 + 12=-2x+18$, or $2x+y=18$.

Answer:

$2x + y=18$